Problem: $h(n) = -5$ $f(t) = -7t-1-5(g(t))$ $g(n) = 3n+7-3(h(n))$ $ g(f(-2)) = {?} $
Explanation: First, let's solve for the value of the inner function, $f(-2)$ . Then we'll know what to plug into the outer function. $f(-2) = (-7)(-2)-1-5(g(-2))$ To solve for the value of $f$ , we need to solve for the value of $g(-2)$ $g(-2) = (3)(-2)+7-3(h(-2))$ To solve for the value of $g$ , we need to solve for the value of $h(-2)$ $h(-2) = -5$ $h(-2) = -5$ That means $g(-2) = (3)(-2)+7+(-3)(-5)$ $g(-2) = 16$ That means $f(-2) = (-7)(-2)-1+(-5)(16)$ $f(-2) = -67$ Now we know that $f(-2) = -67$ . Let's solve for $g(f(-2))$ , which is $g(-67)$ $g(-67) = (3)(-67)+7-3(h(-67))$ To solve for the value of $g$ , we need to solve for the value of $h(-67)$ $h(-67) = -5$ $h(-67) = -5$ That means $g(-67) = (3)(-67)+7+(-3)(-5)$ $g(-67) = -179$